If $z=re^{i\theta}$, how do I prove that $|e^{iz}| = e^{-r \sin(\theta)}$?

Question

If $$ z = re^{i\theta} $$ how do I prove that $$\left| e^{iz} \right| = e^{-r \sin{\theta}} $$

solutions which I cannot understand.

I cannot understand the answer.

Answer 1

Writing $z=re^{i\theta}= r(\cos\theta + i\sin\theta)$, it follows that $$e^{iz} = e^{ir \cos\theta - r\sin\theta} = e^{-r\sin\theta}e^{ir\cos\theta}.$$ Recalling that $\lvert e^{i\alpha} \rvert = 1 \ \forall \alpha \in \mathbb{R}$ and that $e^x > 0 \ \forall x \in \mathbb{R}$, it follows that $$\lvert e^{iz} \rvert = \lvert e^{-r\sin\theta} \rvert \lvert e^{ir\cos\theta} \rvert = e^{-r\sin\theta}.$$

Answer 2

Just change z in the second expresion for its value as $r (\cos \theta + i \sin \theta)$

Now, $e^{ir (\cos \theta + i \sin \theta)} = e^{-r \sin \theta + ir \cos \theta}$ The module of that is just $e^{-r \sin \theta}$ since $e^{ix}$ when $x\in \mathbb{R}$ is just a rotation, with no effects on module.